Integrand size = 11, antiderivative size = 106 \[ \int \tanh ^p\left (a+\frac {\log (x)}{4}\right ) \, dx=e^{-4 a} \left (-1+e^{2 a} \sqrt {x}\right )^{1+p} \left (1+e^{2 a} \sqrt {x}\right )^{1-p}-\frac {2^{1-p} e^{-4 a} p \left (-1+e^{2 a} \sqrt {x}\right )^{1+p} \operatorname {Hypergeometric2F1}\left (p,1+p,2+p,\frac {1}{2} \left (1-e^{2 a} \sqrt {x}\right )\right )}{1+p} \]
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Time = 0.06 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {5652, 383, 81, 71} \[ \int \tanh ^p\left (a+\frac {\log (x)}{4}\right ) \, dx=e^{-4 a} \left (e^{2 a} \sqrt {x}-1\right )^{p+1} \left (e^{2 a} \sqrt {x}+1\right )^{1-p}-\frac {e^{-4 a} 2^{1-p} p \left (e^{2 a} \sqrt {x}-1\right )^{p+1} \operatorname {Hypergeometric2F1}\left (p,p+1,p+2,\frac {1}{2} \left (1-e^{2 a} \sqrt {x}\right )\right )}{p+1} \]
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Rule 71
Rule 81
Rule 383
Rule 5652
Rubi steps \begin{align*} \text {integral}& = \int \left (-1+e^{2 a} \sqrt {x}\right )^p \left (1+e^{2 a} \sqrt {x}\right )^{-p} \, dx \\ & = 2 \text {Subst}\left (\int x \left (-1+e^{2 a} x\right )^p \left (1+e^{2 a} x\right )^{-p} \, dx,x,\sqrt {x}\right ) \\ & = e^{-4 a} \left (-1+e^{2 a} \sqrt {x}\right )^{1+p} \left (1+e^{2 a} \sqrt {x}\right )^{1-p}-\left (2 e^{-2 a} p\right ) \text {Subst}\left (\int \left (-1+e^{2 a} x\right )^p \left (1+e^{2 a} x\right )^{-p} \, dx,x,\sqrt {x}\right ) \\ & = e^{-4 a} \left (-1+e^{2 a} \sqrt {x}\right )^{1+p} \left (1+e^{2 a} \sqrt {x}\right )^{1-p}-\frac {2^{1-p} e^{-4 a} p \left (-1+e^{2 a} \sqrt {x}\right )^{1+p} \operatorname {Hypergeometric2F1}\left (p,1+p,2+p,\frac {1}{2} \left (1-e^{2 a} \sqrt {x}\right )\right )}{1+p} \\ \end{align*}
Time = 1.02 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.14 \[ \int \tanh ^p\left (a+\frac {\log (x)}{4}\right ) \, dx=\frac {e^{-4 a} \left (-1+e^{2 a} \sqrt {x}\right ) \left (\frac {-1+e^{2 a} \sqrt {x}}{2+2 e^{2 a} \sqrt {x}}\right )^p \left (2^p (1+p) \left (1+e^{2 a} \sqrt {x}\right )-2 p \left (1+e^{2 a} \sqrt {x}\right )^p \operatorname {Hypergeometric2F1}\left (p,1+p,2+p,\frac {1}{2}-\frac {1}{2} e^{2 a} \sqrt {x}\right )\right )}{1+p} \]
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\[\int \tanh \left (a +\frac {\ln \left (x \right )}{4}\right )^{p}d x\]
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\[ \int \tanh ^p\left (a+\frac {\log (x)}{4}\right ) \, dx=\int { \tanh \left (a + \frac {1}{4} \, \log \left (x\right )\right )^{p} \,d x } \]
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\[ \int \tanh ^p\left (a+\frac {\log (x)}{4}\right ) \, dx=\int \tanh ^{p}{\left (a + \frac {\log {\left (x \right )}}{4} \right )}\, dx \]
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\[ \int \tanh ^p\left (a+\frac {\log (x)}{4}\right ) \, dx=\int { \tanh \left (a + \frac {1}{4} \, \log \left (x\right )\right )^{p} \,d x } \]
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\[ \int \tanh ^p\left (a+\frac {\log (x)}{4}\right ) \, dx=\int { \tanh \left (a + \frac {1}{4} \, \log \left (x\right )\right )^{p} \,d x } \]
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Timed out. \[ \int \tanh ^p\left (a+\frac {\log (x)}{4}\right ) \, dx=\int {\mathrm {tanh}\left (a+\frac {\ln \left (x\right )}{4}\right )}^p \,d x \]
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